If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?

 Given:

The radius of the base of a cone is halved, keeping the height same.

To do:

We have to find the ratio of the volume of the reduced cone to that of the original cone.

Solution:

Let $r$ be the radius and $h$ be the height of the original cone.

This implies,

Volume of the original cone $=\frac{1}{3} \pi r^2h$

By halving the radius and keeping the height same,

Volume of the new cone $=\frac{1}{3} \pi (\frac{r}{2})^{2} h$

$=\frac{1}{3} \pi \frac{r^{2}}{2} h$

$=\frac{1}{4}(\frac{1}{3} \pi r^{2} h)$

Therefore,

Ratio between the two cones $=\frac{1}{3} \pi r^{2} h: \frac{1}{4}(\frac{1}{3} \pi r^{2} h)$

$=1: \frac{1}{4}$

$=4: 1$

The ratio of the volume of the reduced cone to that of the original cone is $1: 4$.

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Updated on: 10-Oct-2022

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